UNIT-6
Q1:
Solve the following complex function:
Solution:
Given,
Q 2:
Using the residue theorem to evaluate:
Solution:
The singularity z=0,is in our region and we will add the following residue
The singularity z = -3/2 will be skipped because the singularity is not in our region.
The singularity z= -2/3 is in our region and we will add the following residue
Our sum is,
Therefore the solution is,
Q 5:
Prove that the function 
is an analytical function.
Solution. Let 
=u+iv
Let 
=u and 
=v
Hence e-R-Equation satisfied.
Prove that 
Answer Given that
Since 
V=2xy
Now 
But 
Hence 
Q 6:
Find the bi-linear transformation which aps points z=2,1,0 ontpo the points w=1,0,i
Solution:
Let 
Thus we have
= 
Q 7:
How that the bi-liear transformation w= 
transforms 
in the z-plane to 4u+3=0 in w-plane.
Solution:
Consider 
the circle in z-plane
= 0
Thus,centre of the circle is (h,k)
c(2,0) and radius r=2.
Thus in z-plane it is given as 
=2....(1)
Consider w= 
W(z-4) = 2z+3
Wz-4w=2z+3
Wz-2z=4w+3
Z(w-2) = (4w+3)
z = 
z-2 = 
- 2
Q 8:
where C = 
where f(z) = cosz
= 
by cauchy’s integral formula
= 
Q 9:
Solve the following by cauchy’s integral method:
Solution:
Given,
= 
= 
= 
Q 10:
Evaluate the following integral using residue theorem
where c is the circle.
.
Ans. The poles of the integral are given by putting the denominator equal to zero
The integral is analytic on 
and all points inside except
as a pole at 
is inside the circle 
Hence by residue theorem
Q 11:
Evaluate 
where c;|z|=4
Answer Here f(z)=
Poles are 
Sin iz=0
Poles 
Lie inside the circle |z|=4
The given function 
is of the form 
Its pole at z=a is 
Residue (at 
Residue at z=0=
Residue at 
=
Residue at 
are
Respectively -1,1 and -1
Hence the required integrand
Answer 
=
This shows that z=0 is a pole of order 2 for the function 
and the residue of the poles is zero(coefficient of 1/z)
Now the pole at z=0 lies within c
2. Evaluation of definite integrand
Show that 
Solution I= 
Real part of 
Now I= 
=
Putting z=
where c is the unit circle |z|=1
I=
Now f(z) has simple poles at 
and z=-2 of which only 
lies inside c.
Residue at 
is 
=
=
Now equating real parts on both sides we get
I=
3. Prove that 
Solution Let 
Putting 
where c is the unit circle |z|=1
2ai
Poles of f(z) are given by the roots of
Or 
Let 
Clearly 
and since 
we have 
Hence the only pole inside c is at z=
Residue (at 
)
4. Evaluate 
Answer Consider 
Where c is the closed contour consisting of
1) Real axis from 
2) Large semicircle in the upper half plane given by |z|=R
3) The real axis -R to 
and
4) Small semicircle given by |z|=
Now f(z) has simple poles at z=0 
of which only z=
is avoided by indentation
Hence by Cauchy’s Residue theorem
Since 
and
Hence by Jordan’s Lemma 
Also since 
Hence 
Hence as 
Equating imaginary parts we get
Q 12:
Prove that 
Solution Consider 
Where c is the contour consisting of a large semicircle in the upper half plane indented at the origin as shown in the figure
Here we have avoided the branch point o, of 
by indenting the origin
Then only simple of f(z) within c is at z=i
The residue(at z=i) =
Hence by residue theorem
Since 
on -ve real axis.
Now 
Similarly 
Hence when 
Equating real parts we get
